Consider the set $\Omega$ of countable ordinals and let $\omega_1$ be the first uncountable ordinal, $\Omega^*= \Omega \cup \{\omega_1\}$ and equip $\Omega^*$ with the order topology.
I know that $\Omega^*$ is compact Hausdorff. I'm asked to show that
$\Omega$ is an open subset of $\Omega^*$ that is not $\sigma$-compact.
Attempt: Clearly the complement $\Omega^* \setminus \Omega = \{\omega_1\}$ is closed so $\Omega$ is open. Suppose to the contrary that $\Omega$ is $\sigma$-compact (in the relative topology). Then since $\Omega$ itself is locally compact Hausdorff this is equivalent with saying that $\Omega$ is Lindelöf. However, $$\{ \left[0,a\right[ : a < \omega_1\}$$ is an open cover of $\Omega$ consisting of countable sets so $\Omega$ has no countable subcover because $\Omega= \omega_1$ is uncountable and the countable union of countable sets is countable. This is a contradiction. $\quad \square$
Is the above correct?
Best Answer
Every closed subspace of an ordinal space is homeomorphic to an ordinal, and it is compact if and only if it is isomorphic to a successor ordinal.
It follows, therefore, that every compact subspace of $\omega_1$ is countable. Since a countable union of countable sets is countable, $\omega_1$ cannot be $\sigma$-compact.
(Do note that we are using the axiom of choice, as one expects when doing topology.)